Problem: Find the focus of the the parabola $y = x^2.$
Explanation: Recall that a parabola is defined as the set of all points that are equidistant to the focus $F$ and the directrix.

Since the parabola $y = x^2$ is symmetric about the $y$-axis, the focus is at a point of the form $(0,f).$  Let $y = d$ be the equation of the directrix.

[asy]
unitsize(1.5 cm);

pair F, P, Q;

F = (0,1/4);
P = (1,1);
Q = (1,-1/4);

real parab (real x) {
  return(x^2);
}

draw(graph(parab,-1.5,1.5),red);
draw((-1.5,-1/4)--(1.5,-1/4),dashed);
draw(P--F);
draw(P--Q);

dot("$F$", F, NW);
dot("$P$", P, E);
dot("$Q$", Q, S);
[/asy]

Let $(x,x^2)$ be a point on the parabola $y = x^2.$  Then
\[PF^2 = x^2 + (x^2 - f)^2\]and $PQ^2 = (x^2 - d)^2.$  Thus,
\[x^2 + (x^2 - f)^2 = (x^2 - d)^2.\]Expanding, we get
\[x^2 + x^4 - 2fx^2 + f^2 = x^4 - 2dx^2 + d^2.\]Matching coefficients, we get
\begin{align*}
1 - 2f &= -2d, \\
f^2 &= d^2.
\end{align*}From the first equation, $f - d = \frac{1}{2}.$  Since $f^2 = d^2,$ $f = d$ or $f = -d.$  We cannot have $f = d,$ so $f = -d.$  Then $2f = \frac{1}{2},$ so $f = \frac{1}{4}.$

Thus, the focus is $\boxed{\left( 0, \frac{1}{4} \right)}.$